To solve the equation:

\[
x^2 - \frac{1}{x^2} = 12
\]

### Step 1: Multiply the entire equation by \( x^2 \) to eliminate the fraction:

\[
x^4 - 1 = 12x^2
\]

### Step 2: Rearrange the equation:

\[
x^4 - 12x^2 - 1 = 0
\]

Now, let \( y = x^2 \). Substituting this into the equation, we get a quadratic in terms of \( y \):

\[
y^2 - 12y - 1 = 0
\]

### Step 3: Solve the quadratic equation using the quadratic formula:

\[
y = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(-1)}}{2(1)}
\]

Simplifying the terms:

\[
y = \frac{12 \pm \sqrt{144 + 4}}{2} = \frac{12 \pm \sqrt{148}}{2} = \frac{12 \pm 2\sqrt{37}}{2}
\]

\[
y = 6 \pm \sqrt{37}
\]

### Step 4: Back-substitute \( y = x^2 \). So:

\[
x^2 = 6 + \sqrt{37} \quad \text{or} \quad x^2 = 6 - \sqrt{37}
\]

Since \( 6 - \sqrt{37} \) is negative and \( x^2 \) cannot be negative, we discard this solution.

Thus, \( x^2 = 6 + \sqrt{37} \).

### Step 5: Solve for \( x \):

\[
x = \pm \sqrt{6 + \sqrt{37}}
\]

So the solutions are:

\[
x = \sqrt{6 + \sqrt{37}} \quad \text{or} \quad x = -\sqrt{6 + \sqrt{37}}
\]

Would you like further clarification or details on any step?

Here are some related questions to consider:
1. How does the quadratic formula work in more general cases?
2. What is the process for solving more complex equations involving fractions and powers?
3. Can all equations with powers be reduced to a quadratic form?
4. What are the conditions for discarding solutions in equations like this?
5. How can we approximate \( \sqrt{37} \) to get a numerical value for \( x \)?

**Tip**: When solving equations involving fractions and powers, consider substituting variables to simplify the process.

Discover how to solve the equation x^2 - 1/x^2 = 12 through a step-by-step process involving algebraic substitution and the quadratic formula. This solution highlights key algebra concepts suitable for Grades 9-11 students.

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